3. Sodium hydroxide reacts with oxalic acid as shown in the reaction below. H.GO

Question

3. Sodium hydroxide reacts with oxalic acid as shown in the reaction below. H.GO.(s) + 2 NaOH(aq) Na.GOdaq) + 2 H,0(1) a) If 110 mL of 10% (mass per volume) sodium hydroxide is reacted with 4.59 g oxalic acid, calculate the moles of the excess reactant remains after the reaction is complete? Be clear about which species you are calculating. b) This question is unrelated to part(a). The percent yield of the above reaction is found to be 87.6%, if you require 17.5 kg of sodium oxalate for a subsequent reaction, how many kilograms of oxalic acid must be added to excess sodium hydroxide?

Explanation / Answer

a)

mol of acid = mass/MW = 4.59/90.03 = 0.05098

mol of NaOH = mass/MW = (C*V)/MW = (0.1)(110)/(40)= 0.275

2mol of NAOH --> 1 mol of acid

0.05098 mol of acid --> 0.05098*2= 0.10196 mol of base reacted

mol of base left = 0.275-0.10196 = 0.17304 mol of base

b)

mol of NaOx --> mass/MW = 17.5/134 = 0.13059 kmol of Na2C2O4

this is at 87.6% so --> 0.13059/(87.6/100) = 0.14907 kmol of H2C2O4 were used

mass = kmol*MW = 0.14907*90.0349 = 13.421 kg

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